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divider drift: when 5ppm + 5ppm = 5 ppm

Resistor divider drift: when 5ppm + 5ppm = 5 ppm

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Pete Semig

Jun 19 2014 15:17 PM

Comments 3

If you have a voltage divider where each resistor
has 5 ppm/°C drift, what is the worst-case drift?  That is the loaded
question I posed to my colleagues recently (only after figuring out the answer
myself, of course) while working on a low-drift current sensing reference design
(TIPD156).  The ‘obvious’ answer is 10 ppm/°C.  It
turns out to be just 5 ppm/°C, but only when the voltage divider ratio is
½.  Let us delve deeper into the answer to this obvious, yet not-so-obvious
question.

Figure 1 depicts a discrete solution that provides a reference voltage
(VREF) and bias voltage (VBIAS) based on the ratio of
R1 and R2.

Figure 1: Dual reference discrete topology

At the time it was ‘obvious’ that the overall drift of the resistor divider
is (5 ppm/°C) + (5 ppm/°C) = 10 ppm/°C.
Just to be sure, though, I ran a simulation.  Figure 2 depicts the result
of the TINA-TI simulation where R2 drifts in the
positive direction, R1 drifts negatively, and the change in
temperature is 100°C.

Figure 2: Vbias after R1 and R2 drift

Calculating the gain error yields 0.05% (500 ppm or 5.0 ppm/°C), which is the
drift of one resistor…not their sum!

Now let’s examine the familiar voltage divider circuit’s gain ratio and
re-arrange it as shown in Equation ( 1 ).

Where

α is the ratio of R1 and R2, and ultimately relates to
the gain of the circuit.  For example, if α=1, the gain is ½.  So, as
α→0, the gain→1.  Similarly, as α→∞, gain→ 0.

The actual gain of the circuit over temperature will depend on the drift of
the resistors, which is shown in Equation ( 3 ) by δ.  To be
consistent with the simulation, R2 drifts positively and
R1 negatively.

Equation ( 4 ) calculates the gain error.  For simplicity, it is
not converted to a percentage (if you want to do that, simply multiply it by 100
but be careful when converting to ppm).

Using Equations ( 1 ) to ( 3 ), Equation ( 4
) simplifies to:

Notice that the gain error depends on both the drift (δ) and the circuit gain
(related to α).  Furthermore, if α=1 (circuit gain=½), the gain error
simplifies to δ, which is exactly what the simulation shows!

Figure 3 is a plot of the gain error versus the circuit gain for this design
(5 ppm/°C resistor drift, or 500 ppm total drift over 100°C).  Notice
that the gain error is 500 ppm when the circuit gain is ½.  Also notice
that as the circuit gain increases, the gain error decreases and vice versa.

In summary, the drift of the voltage divider was not as expected…it is not
the simple sum and it depends on the circuit gain.  So, next time you’re
designing a circuit and come across one of those ‘obvious’ conclusions, you may
want to double-check it with TINA-TI!

If you have any similar ‘obvious’ experiences, please share them in the
comments section!

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$post.CommentCount)

TINA-TI,
TIPD156,
REF2030

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Comments

EdWalker

Jul 03 2014 08:00 AM

Moral of the story, there isn't much to gain by being a drifter.

Lothar
Kowalik

Jul 07 2014 01:41 AM

Hallo,

there seem to be a simple context when I look at figure 3. This formula is
easyer to remeber:

Gain error = 2 x TK x (1 - division ratio) x delta T

(delta T is the changn in temperatur)  Is the formula right or are
there any argument against it?

Regards from Berlin

Chris
Belting

Jul 07 2014 10:15 AM

It is an interesting observation that I also overlooked for many years
until I used Excel to graph the effect of any tolerance artifact for a voltage
divider. The ratio and location of the two resistors in the divider defines
the extent of the error that can be expected. Assuming both resistors have the
same tolerance, when the resistors are of equal value the maximum error is
equal to the tolerance not the sum of tolerances. When, in a simple divider,
the low-side resistor is significantly less than the high-side resistor the
resulting error approaches the sum of the tolerances. Reverse the relative
locations and the error approaches zero. Plotting the range over 0.001 to 1000
on a log scale is a great way to visualize the point.

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